IPhO 2012 · Theory problem 2

Kelvin water dropper

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Problem T2. Kelvin water dropper (8 points)

The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, \(U=\sigma S\), where \(S\) is the surface area of the interface and \(\sigma\) is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force \(F=\sigma l\), where \(l\) is the length of a straight line separating the fragments.

Figure 1

A long metallic pipe with internal diameter \(d\) is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is \(\sigma\) and its density is \(\rho\). A droplet of radius \(r\) hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration \(g\). Always assume that \(d\ll r\).

Part A. Single pipe (4 points)

i. (1.2 pts) Find the radius \(r_{\max}\) of a drop just before it separates from the nozzle.

ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s electrostatic potential is \(\varphi\). Find the charge \(Q\) of a drop when its radius is \(r\).

iii. (1.6 pts) Consider the situation in which \(r\) is kept constant and \(\varphi\) is slowly increased. The droplet becomes unstable and breaks into pieces if the hydrostatic pressure inside the droplet becomes smaller than the atmospheric pressure. Find the critical potential \(\varphi_{\max}\) at which this will happen.

Part B. Two pipes (4 points)

An apparatus called the “Kelvin water dropper” consists of two pipes, each identical to the one described in Part A, connected via a T-junction, see fig. The ends of both pipes are at the centres of two cylindrical electrodes (with height \(L\) and diameter \(D\) with \(L\gg D\gg r\)). For both tubes, the dripping rate is \(n\) droplets per unit time. Droplets fall from height \(H\) into conductive bowls underneath the nozzles, cross-connected to the electrodes as shown in the diagram. The electrodes are connected via a capacitance \(C\). There is no net charge on the system of bowls and electrodes. Note that the top water container is earthed as shown. The first droplet to fall will have some microscopic charge which will cause an imbalance between the two sides and a small charge separation across the capacitor.

Figure 2

i. (1.2 pts) Express the absolute value of the charge \(Q_0\) of the drops as they separate from the tubes, and at the instant when the capacitor’s charge is \(q\). Express \(Q_0\) in terms of \(r_{\max}\) (from Part A-i) and neglect the effect described in Part A-iii.

ii. (1.5 pts) Find the dependence of \(q\) on time \(t\) by approximating it with a continuous function \(q(t)\) and assuming that \(q(0)=q_0\).

iii. (1.3 pts) The dropper’s functioning can be hindered by the effect shown in Part A-iii. In addition, a limit \(U_{\max}\) to the achievable potential between the electrodes is set by the electrostatic push between a droplet and the bowl beneath it. Find \(U_{\max}\).

Official English text of the problem.

Answers

Here \(\varepsilon_0\) is the vacuum permittivity, and \(q\) denotes the magnitude of the charge on either capacitor plate.

  • A-i: \[ \boxed{r_{\max}=\left(\frac{3\sigma d}{4\rho g}\right)^{1/3}}. \]

  • A-ii: \[ \boxed{Q=4\pi\varepsilon_0r\varphi}. \]

  • A-iii: \[ \boxed{\varphi_{\max}=2\sqrt{\frac{\sigma r}{\varepsilon_0}}}. \] The same threshold applies to the magnitude of a negative potential.

  • B-i: \[ \boxed{Q_0=\frac{2\pi\varepsilon_0r_{\max}}{C}\,q}. \]

  • B-ii: In the continuous approximation with negligible flight delay, \[ \boxed{q(t)=q_0\exp\!\left(\frac{2\pi\varepsilon_0nr_{\max}}{C}\,t\right)}. \]

  • B-iii: The limit due to electrostatic repulsion is \[ \boxed{ U_{\max} =r_{\max}\sqrt{\frac{2\rho gH}{3\varepsilon_0}} =\sqrt{\frac{\sigma dH}{2\varepsilon_0r_{\max}}} }. \]

No numerical parameter values are supplied.

Solution

Part A-i

The slowly growing drop is in mechanical equilibrium until its weight exceeds the maximum supporting force of surface tension.

The nozzle circumference is \(\pi d\). When the surface tension acts vertically around the neck, its maximum upward force is \[ F_\sigma=\pi d\sigma. \] The neck volume and the pressure-force correction at the small nozzle opening are negligible for \(d\ll r\). Approximating the drop as spherical, its weight is \[ mg=\frac{4}{3}\pi r^3\rho g. \]

At detachment, \[ \frac{4}{3}\pi r_{\max}^3\rho g=\pi d\sigma, \] hence \[ \boxed{r_{\max}=\left(\frac{3\sigma d}{4\rho g}\right)^{1/3}}. \]

The spherical approximation is consistent with this result: the ratio of gravitational to capillary stresses is \[ \frac{\rho g r_{\max}^2}{\sigma} =\frac{3d}{4r_{\max}}\ll1. \]

Part A-ii

The conducting drop and pipe have the same potential. Neglecting the thin pipe’s perturbation to the drop’s electric field, the drop behaves as a conducting sphere.

Outside a sphere carrying charge \(Q\), Gauss’s law gives \[ E(s)=\frac{Q}{4\pi\varepsilon_0s^2}. \] Taking the distant surroundings to have zero potential, \[ \varphi=\int_r^\infty E(s)\,ds =\frac{Q}{4\pi\varepsilon_0r}. \] Therefore \[ \boxed{Q=4\pi\varepsilon_0r\varphi}. \]

The sign of the drop’s charge is the sign of \(\varphi\).

Part A-iii

Surface tension produces an inward capillary pressure. From the given surface energy, \[ W_\sigma=4\pi r^2\sigma, \] the pressure required to increase the surface area is \[ p_\sigma=\frac{dW_\sigma}{dV} =\frac{8\pi r\sigma}{4\pi r^2} =\frac{2\sigma}{r}. \]

The electric field produces an outward pressure. For a conductor, the surface charge density is \(\varepsilon_0E\), and the field acting on that charge due to the other charges is \(E/2\). Thus \[ p_{\mathrm e}=\frac{\varepsilon_0E^2}{2}. \] Using Part A-ii, \[ E=\frac{Q}{4\pi\varepsilon_0r^2}=\frac{\varphi}{r}. \]

Consequently, mechanical equilibrium gives \[ p_{\mathrm{in}}-p_{\mathrm{atm}} =p_\sigma-p_{\mathrm e} =\frac{2\sigma}{r} -\frac{\varepsilon_0\varphi^2}{2r^2}. \]

According to the specified instability criterion, the threshold occurs when the internal pressure equals atmospheric pressure: \[ \frac{2\sigma}{r} =\frac{\varepsilon_0\varphi_{\max}^2}{2r^2}. \] Therefore \[ \boxed{\varphi_{\max}=2\sqrt{\frac{\sigma r}{\varepsilon_0}}}. \]

Equivalently, the corresponding charge magnitude is \[ Q_{\mathrm{crit}} =4\pi\varepsilon_0r\varphi_{\max} =8\pi\sqrt{\varepsilon_0\sigma r^3}. \]

Part B-i

Write \[ R=r_{\max}, \qquad U=\frac{q}{C}. \] We neglect stray capacitances compared with \(C\). By symmetry and zero net charge, the two electrode potentials relative to earth are \[ V_{\mathrm L}=+\frac{U}{2}, \qquad V_{\mathrm R}=-\frac{U}{2}. \]

Since \(L\gg D\), the potential near the centre of each cylinder, before accounting for the small drop, is approximately the cylinder’s potential. Since \(D\gg R\), the drop’s own contribution to its surface potential is approximately that of an isolated sphere.

The water and pipes are grounded. Thus, while the left drop remains attached, \[ 0=V_{\mathrm L}+\frac{Q_{\mathrm L}}{4\pi\varepsilon_0R}. \] It follows that \[ Q_{\mathrm L}=-2\pi\varepsilon_0RU. \] Similarly, \[ Q_{\mathrm R}=+2\pi\varepsilon_0RU. \]

The charge is retained upon separation, so its magnitude is \[ \boxed{ Q_0=2\pi\varepsilon_0RU =\frac{2\pi\varepsilon_0r_{\max}}{C}\,q }. \]

Because of the cross-connections, each drop falls into a bowl already carrying charge of the same sign as the drop. Its arrival therefore reinforces the charge separation.

Part B-ii

Use the continuous approximation and neglect the delay between detachment and collection. This assumes that the voltage changes little during a drop’s flight.

Each tube produces \(n\) drops per unit time. The positive capacitor plate receives positive charge at rate \(nQ_0\), while the negative plate receives negative charge at the same rate. Hence the magnitude \(q\) on either plate obeys \[ \frac{dq}{dt}=nQ_0 =\frac{2\pi\varepsilon_0nR}{C}\,q. \] There is no additional factor of two: a pair of drops increases the plate charges from \(+q,-q\) to \(+(q+Q_0),-(q+Q_0)\).

Integrating and imposing \(q(0)=q_0\), \[ \ln\frac{q(t)}{q_0} =\frac{2\pi\varepsilon_0nR}{C}\,t, \] so \[ \boxed{ q(t)=q_0\exp\!\left(\frac{2\pi\varepsilon_0nr_{\max}}{C}\,t\right) }. \]

The characteristic growth time is \[ \tau=\frac{C}{2\pi\varepsilon_0nr_{\max}}. \] The negligible-delay approximation requires the flight time to be small compared with \(\tau\). The exponential growth applies before breakup or electrostatic repulsion significantly interferes with collection.

Part B-iii

Consider the drop formed inside the positive electrode. It carries charge \(-Q_0\), and its collecting bowl is connected to the negative electrode.

Treat \(U\) as approximately constant during the fall. The potential produced by the apparatus changes from \(+U/2\) near the nozzle to \(-U/2\) at the bowl. Therefore the increase in the drop’s external electrostatic potential energy is \[ \Delta W_{\mathrm e} =(-Q_0)\left(-\frac{U}{2}-\frac{U}{2}\right) =Q_0U. \]

The potential used here is the potential produced by the apparatus. The drop’s zero total potential while attached also includes its own field. During flight, its charge and radius remain approximately constant, so its self-energy does not contribute to the translational work.

Neglecting air resistance and initial kinetic energy, the gravitational energy available is \[ mgH=\frac{4}{3}\pi R^3\rho gH. \] For collection to remain possible, this must be sufficient to overcome the electrostatic energy increase: \[ mgH\ge Q_0U. \]

At the energy limit, using \(Q_0=2\pi\varepsilon_0RU\), \[ \frac{4}{3}\pi R^3\rho gH =2\pi\varepsilon_0R\,U_{\max}^2. \] Thus \[ \boxed{ U_{\max}=r_{\max}\sqrt{\frac{2\rho gH}{3\varepsilon_0}} }. \]

Using the detachment condition \(mg=\pi d\sigma\), this can also be written \[ \boxed{ U_{\max}=\sqrt{\frac{\sigma dH}{2\varepsilon_0r_{\max}}} }. \]

The separate breakup threshold from Part A-iii, evaluated for a drop of radius \(r_{\max}\), corresponds to \[ U_{\mathrm{break}} =2\varphi_{\max}(r_{\max}) =4\sqrt{\frac{\sigma r_{\max}}{\varepsilon_0}}. \] This provides an additional restriction on operation with intact drops.

The solution was checked against the official solution.